Question: In the complex plane, the graph of $|z - 3| = 2|z + 3|$ intersects the graph of $|z| = k$ in exactly one point.  Find all possible values of $k.$

Enter all possible values, separated by commas.
Answer: Let $z = x + yi,$ where $x$ and $y$ are real numbers.  Then the equation $|z - 3| = 2|z + 3|$ becomes
\[|x + yi - 3| = 2 |x + yi + 3|,\]so
\[(x - 3)^2 + y^2 = 4[(x + 3)^2 + y^2].\]This simplifies to $x^2 + 10x + y^2 + 9 = 0.$  Completing the square, we get
\[(x + 5)^2 + y^2 = 4^2.\]This is the circle centered at $-5$ with radius 4.

[asy]
unitsize(0.4 cm);

draw(Circle((0,0),1),red);
draw(Circle((0,0),9),red);

draw(Circle((-5,0),4));
draw((-10.5,0)--(10.5,0));
draw((0,-10.5)--(0,10.5));

label("$4$", (-3,0), N);
label("$4$", (-7,0), N);

dot("$-5$", (-5,0), S);
[/asy]

The graph of $|z| = k$ is a circle centered at the origin with radius $k.$  We see that the circles with radius $\boxed{1}$ and $\boxed{9}$ intersect the circle $(x + 5)^2 + y^2 = 4^2$ in exactly one point.